$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$

I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.


The above method is really verifying and always quick. Another method to arrive at the answer is by rationalising denominator (mainly when the answer [or RHS] is not known or one is asked to work out only from LHS to RHS):

$$\frac{\sin x - \cos x + 1 }{\sin x + \cos x - 1 }\cdot \frac{\sin x + \cos x + 1}{\sin x + \cos x + 1}$$

$$\frac{ (\sin x + 1)^2 - \cos^2 x }{ 2 \sin x \cos x } $$

$$ \frac{ \sin^2 x + 2 \sin x + 1 - \cos^2 x }{ 2 \sin x \cos x } $$

$$ \frac{ \sin^2 x + 2 \sin x + \sin^2 x + \cos^2 x - \cos^2 x } {2 \sin x \cos x } $$

and the answer follows i.e. $$ \frac{\sin x + 1}{\cos x}. $$

Hope it was helpful.

  • $\begingroup$ well done i say :) $\endgroup$ – user87543 Dec 12 '13 at 13:47
  • 1
    $\begingroup$ I wouldn't call this rationalizing the denominator, as there's nothing necessarily rational or irrational about the denominator before or after the initial multiplication, but this technique does mirror the technique for rationalizing denominators: using conjugates. $\endgroup$ – Isaac Dec 12 '13 at 16:05


$$\frac{a}{b}=\frac c d\iff ad=bc$$

  • $\begingroup$ and obviously this is allowed when $(b,d)\not=0$ which is when $x \not=\pi/2 + k\pi$ and $x\not=k\pi$ $\endgroup$ – Jekyll Dec 12 '13 at 13:39
  • $\begingroup$ Exactly what's needed here! +1 $\endgroup$ – Namaste Dec 13 '13 at 16:00

$(\sin x- \cos x+1)\cos x = \sin x \cos x -\cos^2 x +\cos x$

$(\sin x+ \cos x-1)(\sin x +1) = \sin^2 x + \sin x \cos x +\cos x -1 = \sin x \cos x +\cos x +(\sin^2 x -1)= \sin x \cos x -\cos^2 x +\cos x$


As always, the method that "always" (never say "never" OR "always"...) work is the substitution $ t = \tan \frac{x}{2},$ which makes $\sin(x) = \frac{2x}{1+x^2}, \cos(x)= \frac{1-x^2}{1+x^2},$ which makes an identity like this a mechanical verification.


Observe that the Right Hand side $\displaystyle\frac{\sin x+1}{\cos x}=\tan x+\sec x$

So, I want to utilize $\displaystyle\sec^2x-\tan^2x=1$

Dividing the numerator & the denominator by $\cos x,$

$$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\tan x-1+\sec x}{\tan x+1-\sec x}$$

$$=\frac{\tan x+\sec x-(\sec^2x-\tan^2x)}{\tan x+1-\sec x}(\text{ Replacing }1\text{ with } \sec^2x-\tan^2x)$$

$$=\frac{(\sec x+\tan x)-(\sec x+\tan x)(\sec x-\tan x)}{\tan x+1-\sec x}$$

$$=\frac{(\sec x+\tan x)(1-\sec x+\tan x)}{\tan x+1-\sec x}$$

$$=\sec x+\tan x$$

  • $\begingroup$ @dona12, how about this? $\endgroup$ – lab bhattacharjee Dec 12 '13 at 14:56

I will start like Timotej, but finish differently.

$$\cos x(\sin x-\cos x+1)=\cos x(1+\sin x)-\cos^2x=\cos x(1+\sin x)-(1-\sin^2x)$$ $$=(1+\sin x)\{\cos x-(1-\sin x)\}$$

$$\implies \cos x(\sin x-\cos x+1)=(1+\sin x)(\sin x+\cos x-1)$$

Now change the sides of $\displaystyle \cos x, \sin x+\cos x-1$

Let me derive some other identities

$(1)\displaystyle\sin x(\sin x-\cos x+1)=\sin^2x+\sin x(1-\cos x)=1-\cos^2x+\sin x(1-\cos x)$

$\displaystyle\implies\sin x(\sin x-\cos x+1)=(1-\cos x)(\sin x+\cos x+1)$

$(2)\displaystyle\sin x(\sin x+\cos x-1)=\sin^2x-\sin x(1-\cos x)=(1-\cos x)(1+\cos x-\sin x)$

$(3)\displaystyle\cos x(\sin x+\cos x-1)=\cos^2x-\cos x(1-\sin x)=(1-\sin x)(1+\sin x-\cos x)$

and so on

  • $\begingroup$ @dona12, another method $\endgroup$ – lab bhattacharjee Dec 13 '13 at 3:50

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